Module in stat

8/7/2019 Module in stat

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LESSON 1 : POPULATION AND SAMPLE

I. Specific Objective

After studying this lesson you should be able to:

1. Apply the definition of Population and Sample2. Apply the formula for the Population size and Sample size3. Solve problems involving Population and Sample size

II. Learning Activities

Input

Population Refers to the total of all objects under study

Sample Small part of the population that serves as a representation

Where:

Example 1:Find the sample size of the population is 250 at 95% accuracy

At 95% accuracy, the corresponding percentage error is 5% or .05.

1+ 250(.05)

n = 153.85n = 154

n = N

1+ Ne

N is the Population sizee is the margin of error

n is the Sample size

n = 250


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III. Problem Solving

a. N = 290 ; e = 20%b. N = 521 ; e = 5%c. N = 3520 ; e = 2%d. N = 90 ; e = 10%


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LESSON 2 : PERCENTAGE



1. Apply the definition of Percentage2. Apply the formula for the Percentage3. Solve problems involving Percentage


Input

Percentage Is the number of hundredth part of one number is up another

X 100bo

WhereP% = number of p-arts of 100b1 = data to be compared with the base

bo = is the base use for comparison

Example:Out of 28,000 took the I.I entrance examination, 6,000 accepted in to the institute.What is the percentage of students who were accepted at I.I

Given:b1 = 6,000

X 100 21.43bo = 28,000 28000

What is the percentage marked up if Angie bought a can in the amount

of 8,250.00 and sale it for 9,000.00

Given:b1 = 8,250.00

X 100 91.67Bo = 9,000.00 9000

P% = b1

P% = 6,000

P% = 8,250


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a. b1 = 500 ; bo = 700b. b1 = 25 ; bo = 76c. b1 = 3520 ; bo = 5000d. b1 = 100 ; bo = 900


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cx = cx1 + cx2 + cx3 + . . . . .Cn or c(x1 + x2 + x3 + . . . . .xn)I = 1

n n

cxi = cxiI = 1 I = 1

4 4Example Suppose the value of xi = 16 then evaluate the value 5xi

I = 1 I = 1

4 4

xi = 5 xi = 5 (16) = 80I = 1 I = 1

Consider the ff. Observation x1 = 12, x2 = 7, x3 = 10, x4 = 13, x5 = 8.Determine the value of the ff.

a. 5 b. 5xi 3xiI = 1 I = 1

a. 5

xi = (12 + 7 + 10 + 13 + 8) = 50I = 1

b. 5

3xi = 3 (12 + 7 + 10 + 13 + 8) = 150

I = 1

Rule 3 1f xi and yi are two random quantitive then

n

(xi +yi) = (x1 + y1) + (x2 + y2) + . . . . .(xn + yn)I = 1 or = ( x1 +x2 + . . . xn) + (y1 + y2 + . . . yn)

n n n

(xi +yi) = xi + yiI = 1 I = 1 I = 1

Example x1 = 3, x2 = 5, x3 = 7, x4 = 6y1 = 5, y2 = 8, y3 = 8, y4 = 7

4Evaluate the expression (xi +yi)

I = 1


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4

(xi +yi) = ( 3 + 5) + (5 +8) + (7 + 8) + (6 + 9)I = 1 = 8 + 13 + 15 + 15

= 51

4 (xi +yi) = (3 + 5 + 7 + 6) + ( 5 + 8 + 8 + 9)

I = 1 = 21 + 30= 51


Evaluate the following:

Given: x1 = 12, x2 = 8, x3 = 14, x4 = 6, x5 = 10y1 = 9, y2 = 15, y3 = 11, y4 = 20

5 51 xi 2 4xi

I = 1 I = 1

4 53 (3xi yi) 4

I = 1 I = 3

5 4

I = 1

Consider the ff. Measurements

Given: X = 12. 15, 8, 3 , 7, 10, 15, 15Y = 11, 12, 10, 15, 9, 8, 17, 14

Determine the value of the ff.1 xy 3 52 4 3 ( x + y)

xi

xiyi

xy x xy


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LESSON 4 : FREQUENCY DISTRIBUTION



1. Apply the definition of Frequency Distribution2. Apply the contraction of Frequency Distribution3. Apply the formula of Frequency Distribution4. Solve problems involving Frequency Distribution


Input

Frequency Distribution Data that are summarize into classes on category to

show the occurrence value or objects in each class or

categories

Example

Test Score Obtained by Sixty Students in a STAT class

48 73 57 57 69 88 11 80 87 4746 70 49 45 75 81 33 65 38 5994 59 62 36 58 69 45 55 58 65

30 49 73 29 41 53 37 35 61 4822 51 56 55 60 37 56 59 57 3612 36 50 63 68 30 56 70 53 28

Exam Score F.D. Frequency Distribution11 - 22 3 - 14 323 - 34 5 - 16 535 - 46 11 - 22 1147 - 58 19 - 30 1959 - 70 14 - 25 1471 - 82 6 - 17 683 - 94 2 - 13 2

100 111 100

Construction of Frequency Distribution1 Get the lowest and highest score2 Get the value of the range

R = H L3 Determine the no. of Class

K = 1 + 33 log n4 Determine the size of the class interval

cRK

5 Construct the classes


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LESSON 5: COMULATATIVE FREQUENCYDISTRIBUTION



2. Apply the Greater than Cumulatative Frequency Distribution3. Apply the Related Frequency Distribution2. Apply the Less than Cumulatative Frequency Distribution


Input

%f = f x 100n

Where Greater than Cumulatative Frequency = > cumf

Less than Cumulatative Frequency = < cumf

Related Frequency Distribution = %f

Example

Test Score Obtained by Sixty Students in a STAT class48 73 57 57 69 88 11 80 87 4746 70 49 45 75 81 33 65 38 5994 59 62 36 58 69 45 55 58 6530 49 73 29 41 53 37 35 61 4822 51 56 55 60 37 56 59 57 3612 36 50 63 68 30 56 70 53 28

Exam Score > cumf fd %f


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LESSON 6: MEAN



1. Apply the definition of Mean2. Apply the formula for the Mean ungroup data3. Apply the formula for the Weighted Mean4. Apply the formula for the Mean group data

4.1 Apply the formula of Midpoint data

4.2 Apply the formula of Unit Deviation5. Solve problems involving Mean


Input

Mean - The simplest and most efficient in measuring the central tendency

MEAN FOR UNGROUP DATA

Where

X =Sum of all value in the distributionno. of values of distribution

X =nn

ExampleConsider the following value12, 15, 16, 13, 11, 10, 9, 14, 20

X = 12 + 15 + 16 + 13 + 11 + 10 + 9 + 14 + 20 X = 120 = 13.339 9

ExampleConsider the following value23, 10, 14, 16, 18, 27, 31, 25, 12

Find the X

X =23 + 10 + 14 + 16 + 18 + 27 + 31 + 25 + 12

X =176

= 19.569 9


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WEIGHTED MEAN

Where

X =Sum of the product per unit qtyTotal no. of unit

X =wx wherew x = represent the qty

w = represents the unit associated

ExampleConsiderd we are interest in computing the weighted mean grade of student

Subject No. of Unit Grade wx(w) (x)

1 3 2.00 62 3 3.00 93 5 1.25 6.254 1 3.00 35 2 2.50 56 3 2.50 7.5

w = 17 wx = 36.75

X = wx ### = 2.16w 17

X =3(2) + 3(3) + 5(1.25) + 1(3) + 2(2.50) + 3(2.50

=36.75

17 17

X = 2.16

MEAN FOR GROUP DATAMidpoint Method

X =fxn

wheref = frequency of each datax = for midpoint of each datan = the total no. of frequency

Step1. Get the midpoint of each class2. multiply each midpoint by its coresding frequency3. Get the sum of the product in step 2

4. Devide the sum obtained in step 3 by the total no. of frequency


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Test Score Obtained by Sixty Students in a STAT classClass f x fx

11 - 22 3 16.5 49.523 - 34 5 28.5 142.535 - 46 11 40.5 445.547 - 58 19 52.5 997.559 - 70 14 64.5 90371 - 82 6 76.5 45983 - 94 2 88.5 177

n = 60 fx = 3174

X = fx 3174 = 52.9n 60

Unit Devotion Method

X = Xa+ ( fdnWhere

xa = assumed meanf = frequencyd = distribution

n = sample sizsec = size of the class interval

Example 160 students in Stat Class

Class f d fd11 - 22 3 -3 -9 4723 - 34 5 -2 -10 + 5835 - 46 11 -1 -11 125 / 2= 52.547 - 58 19 0 059 - 70 14 1 1471 - 82 6 2 12

83 - 94 2 3 6n = 60 fd = 2

X = Xa+ ( fd X = 52.5 +( 2n 60= 52.5 +( 260

X = 52.9

Example 2Ages of 70 Mayor

Class f d fd

25 - 30 3 -3 -9 43

)c

)c )12)12


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LESSON 7: MEDIAN



1. Apply the formula for the Median ungroup data2. Apply the formula for the Median group data3. Solve problems involving Median


InputMedian for Ungroup data

X =x(n+1)

odd2

n ) + n evenX =

2 22

ExampleFind the mean of the Following

1 21, 10, 36, 42, 39, 52, 30, 25, 26

To get the mean arrange first the given numbers fromlowest to highest no. and find the odd no.

10, 21, 25, 26, 30, 36, 39, 42, 54

X =

x(9+1)

2

=10

= 5 = 362

Median for Group datan

- cumfb 2

y( x( +1)


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fm

Wherexlb = refers to the lowest boundaries of the median classcumfb = cumulative frequency before the median class

fm = frequency of the median class

ExampleClasses f


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c median


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LESSON 8: MODE



1. to apply the definition of Mode1. Apply the used of Mode1. Apply the formula for the Mode for group data2. Solve problems involving Mode


Input

Mode for Group Data

X = xlb + ( d1d1 - d2Where

d1 = modal class frequency of the interval prociding the modal class

d2 = modal class frequency of the internal after the modal classxlb = lower bounderyMoidal class - hihest frequency of the distribution

ExampleClasses f

70 - 75 276 - 81 882 - 87 1988 - 93 2194 - 99 28 d1

100 - 105 38 Modal class

106 - 111 15 d2

112 - 117 9Given

d1 = 28d2 = 15

X = xlb + ( d1d1 - d2

X = 99.528

28 - 15

)c

)c

+( )6


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= 112.42


Class f 73 - 75 276 - 78 679 - 81 1282 - 84 1685 - 87 1888 - 90 3991 - 93 3694 - 96 2197 - 99 5

Find the mean, median, mode for group data


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LESSON 9: QUARTILES, DECILES,PERCENTILES



1. Apply the parts of each distribution2. Apply the formula for the Quartiles, Deciles and Percentiles3. Solve problems involving Quartiles, Deciles, and Percentiles


Input

(kn

= cumfbQk = xlb + 4

fqkwhere

xlb = lower boundery of the Kth quartile

cumfb = cumulative frequency before the Kth quartilefqk = frequency before the Kth quartile

where

Quartiles = 4 parts of distributionDeciles = 10Percentiles = 100

QUARTILES

Determine the quartile class

Q1 =1

(n + 1)4

Q3 =3

(n + 1)4

Classes f cumf 11 - 22 3 3 cumfb

23 - 34 5 1srt q1 8 1st quartile

35 - 46 11 1947 - 58 19 3859 - 70 14 52 cumfb

71 - 82 6 3rdt q3 58 3rd quartile

83 - 94 2 6060

Q1 =1 (n +1)4

)c


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LESSON 10: AVERAGE DEVIATION



1. Apply the definition of Average Deviation2. Apply the formula for the Average Deviation

2.1 Apply the formula for Ungroup data2.2 Apply the formula for Group data


Input

Average Deviation - Refers to the atentic mean of the absolute deviation

of the values from the mean of the devation

AD = / X - X/

nAverage Devation for Ungroup Data

WhereX = midpoint of each classx = mean of the distributionn = Total no. of frequency

X /X - x/6 - 11 = -5 57 - 11 = -4 49 - 11 = -2 2

AD = / X - X/ 22

11 - 11 = 0 0 n 713 - 11 = 2 2

15 - 11 = 4 4 = 3.1416 - 11 = 5 5

X= 77= 11

227

Average Devotion Group Data

Wheref = frequency of each classX = midpoint of each classx = mean of the distribution

n = Total no. of frequency


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AD = f/ X - X/

n

Classes f X fX11 - 22 3 16.5 49.523 - 34 5 28.5 142.535 - 46 11 40.5 445.5

AD = f/ X - X/

47 - 58 19 52.5 997.5 n59 - 70 14 64.5 903

=3174

71 - 82 6 76.5 459 6083 - 94 2 88.5 177

n = 60 fX= 3174 = 52.9


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LESSON 11: VARIANCE



1. Apply the definition of Variance2. Apply the formula for the Variance

2.1 Apply the formula for Ungroup data2.2 Apply the formula for Group data


Input

Variance - Square of Average Devation

S = (X - x)

nVariance for Ungroup Data

WhereX = represent the individual values in the distributionx = The mean in the listn = is the simple size

Compute the value of the variance of the ff. information13, 5, 7, 9, 10, 17, 15, 12

X X - x (X - x)5 - -6 367 - -4 16

9 - -2 4 S (X - x)10 - -1 1 n12 - 1 113 - 2 4 = 14.2515 - 4 1617 6 36

X= 88= 11

1148

Variance for Group Data

Where


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LESSON 12: STANDARD DEVIATION



2. Apply the formula for the Standard Deviation.2.1 Apply the formula for Ungroup data2.2 Apply the formula for Group data


Input

Standard Deviation - Square root of Variance

Sd = (X - x)

nStandard Deviation for Ungroup Data

WhereX = represent the individual values in the distributionx = The mean in the listn = is the simple size

Compute the value of the variance of the ff. information13, 5, 7, 9, 10, 17, 15, 12

X X - x (X - x)5 - -6 367 - -4 16

9 - -2 4 S (X - x)10 - -1 1 n12 - 1 113 - 2 4 = 14.2515 - 4 1617 6 36 = 14.25

X= 88= 11

114 = 3.778

Variance for Group Data

Where


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